Rảnh rỗi thật sự .-.

a) \(\left(12-12\dfrac{1}{3}\right):x+\dfrac{1}{6}=-\dfrac{2}{3}\)

\(-\dfrac{1}{3}x=-\dfrac{2}{3}-\dfrac{1}{6}\)

\(-\dfrac{1}{3}x=-\dfrac{5}{6}\)

\(x=-\dfrac{5}{6}:\left(-\dfrac{1}{3}\right)\)

\(x=\dfrac{5}{2}\)

b) \(\dfrac{4}{x}=\dfrac{x}{16}\)

\(x^2=4.16\)

\(x^2=64\)

\(\Rightarrow x=8;x=-8\)

`a)=>(12-37/3):x+1/6=-2/3`

`=>(12-37/3):x=-5/6`

`=>(-1/3):x=-5/6`

`=>x=(-1/3):(-5/6)`

`=>x=6/15=2/5`

`b)4/x=x/16`

`=>x^2=4*16`

`=>x^2=64`

`=>x^2=(+-8)^2`

a) Sửa đề: \(A=\left(3x-2\right)\left(9x^2+6x+4\right)-3x\left(9x^2-2\right)\)

\(=27x^3-8-27x^3+6=-2\)

b: Ta có: \(B=\left(3x+5\right)^2+\left(6x+10\right)\left(2-3x\right)+\left(2-3x\right)^2\)

\(=\left(3x+5+2-3x\right)^2\)

=49

\(A=27x^3-8-27x^3+6=-2\)

a: \(\Leftrightarrow8x\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{0;3;-3\right\}\)

b: \(\Leftrightarrow x^2-4x+4-x^2-2x+3=12\)

=>-6x=5

hay x=-5/6

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

c) Ta có: \(\left|x-\dfrac{2}{3}\right|+2.25=\dfrac{3}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{9}{4}=\dfrac{-3}{2}\)(vô lý)

Vậy: \(x\in\varnothing\)

a) Ta có: \(x+\dfrac{-3}{7}=\dfrac{4}{7}\)

\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{4}{7}\)

hay x=1

Vậy: x=1

b) Ta có: \(\dfrac{1}{2}x-75\%=\dfrac{1}{4}\)

\(\Leftrightarrow x\cdot\dfrac{1}{2}=1\)

hay x=2

Vậy: x=2

\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)

a) \(A=\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)

\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)

b) \(B=\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)

\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)

\(=x^4-16-x^4+9=-7\)

a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)